Experiment 7: Comparison of Unknowns: Weak Acid Identification
Introduction: This is a guided inquiry lab that will have seven parts. The main goal is to have a better
understanding of acid-base chemistry. This is done by following guidelines and answer questions
during the lab to get to the final step.
Procedure:
Standardize 0.1M NaOH using KHP. Weigh out about 0.5106g KHP for a 25mL titration. Calibrate
the pH meter and then fill three flasks each with the appropriate amount of KHP and 50mL of water.
Then add the NaOH to the flask of KHP in increments of 2-3mL at a time while take pH readings after
each addition. Once you notice a dramatic change in pH slow the addition to only 0.5mL and do this until
the change slows back down and then you can go back to adding around 2-3mL at once until there is
minimal change. Do this technique until you have three good trials. Once you obtain a dry sample of
your unknown do the titration in the same manor. This lab was done by following the seven steps
handed out in lab and answering the questions during this process.
Data:
KHP Titration
Trial
Eq. Point (mL)
Half Eq. Point (mL)
NaOH (M)
1
25.1mL
12.55mL
0.1000
2
25.1mL
12.55mL
0.1000
3
25.1mL
12.55mL
0.1001
Average
25.1mL
12.55mL
0.1000
Standardizing graphs (3 trials)
Unknown C Titration
Trial
Mass of Unk C
(g)
Volume NaOH
Volume NaOH
added at Eq.
added at Half
Point (mL)
Eq Point (mL)
MW Acid
(g/mol)
pKa
Ka
1
0.3072g
18.2
9.1
168.791
3.9
1.26 ∗ 10−4
2
0.3078g
18.1
9.05
170.055
4.1
7.9 ∗ 10−5
3
0.3076. g
18.0
9
170.889
4.0
1.00 ∗ 10−4
18.1
9.05
169.912
4.0
1.02 ∗ 10−4
Average
Unknown Acid C titrations (3 trials)
Calculations:
Mass of KHP needed
1𝑚𝑜𝑙𝐾𝐻𝑃
204.23𝑔
𝑉𝑁𝑎𝑂𝐻 ∗ 𝑀 ∗
∗
= 𝑔𝐾𝐻𝑃
1𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1𝑚𝑜𝑙 𝐾𝐻𝑃
Molarity of NaOH
1 mol KHP 1 mol NaOH 1 mol NaOH
g KHP x
x
x
= M NaOH
204.23g
1 mol KHP
0.0251 L
Moles of Unknown Acid
1 𝑚𝑜𝑙 𝑢𝑛𝑘 𝑎𝑐𝑖𝑑
𝑉 𝑁𝑎𝑂𝐻 ∗ 𝑀 ∗
= 𝑚𝑜𝑙 𝑢𝑛𝑘 𝑎𝑐𝑖𝑑
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
MW of Unknown Acid
𝑔 𝑈𝑛𝑘𝑜𝑤𝑛 𝑎𝑐𝑖𝑑
= 𝑀𝑊
𝑚𝑜𝑙 𝑢𝑛𝑘𝑜𝑤𝑛 𝑎𝑐𝑖𝑑
pKa of Unknown Acid C
1
𝑝𝑘𝑎 = 𝑝𝐻 𝑎𝑡 𝑉𝑒
2
Ka of Unknown Acid C
𝐾𝑎 = 10−𝑝𝑘𝑎
Mass of KHP needed Example
0.025𝐿 ∗ 0.1𝑀 ∗
1𝑚𝑜𝑙𝐾𝐻𝑃
204.23𝑔
∗
= 0.5106𝑔𝐾𝐻𝑃
1𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1𝑚𝑜𝑙 𝐾𝐻𝑃
Molarity of NaOH trial 1
0.5126g KHP x
1 mol
1 mol
1 mol
x
x
= 0.1000M NaOH
204.23g 1 mol 0.0251 L
Moles of Unknown Acid trial 1
1 𝑚𝑜𝑙 𝑢𝑛𝑘 𝑎𝑐𝑖𝑑
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= .00182𝑚𝑜𝑙
MW of Unknown Acid Trial 1
0.3072𝑔 𝑈𝑛𝑘𝑜𝑤𝑛 𝑎𝑐𝑖𝑑
= 168. 791𝑀𝑊
0.00182𝑚𝑜𝑙 𝑢𝑛𝑘𝑜𝑤𝑛 𝑎𝑐𝑖𝑑
pKa of Unknown Acid C Trial 1
3.9 = 3.9
. 0182𝐿 𝑁𝑎𝑂𝐻 ∗ 0.100𝑀 ∗
Ka of Unknown Acid C Trial 1
1.26 ∗ 10−4 = 10−3.9
Possible Identities of Unknown Acid C:
-1-Naphthoic Acid: 172.18g/mol and Ka=2.1x10-4
-Acetylsalicylic Acid: 180.157g/mol and Ka=3.3x10-4
When looking at our data the formula weight is very close to 1-Naphthoic Acid but the pka we found, 4,
is a little higher than the expected 3.68. This could be due to estimating with the graph and not getting
an exact equivalence point and not getting an exact half equivalence point. I probably should have point
more points on the graph and printed it out in order to make it more accurate. For Acetylsalicylic acid
our formula weight and pKa and Ka are even further off from the expected values. Also another error
could have been with the pH meter and not waiting for it to stabilize so my pHs could be off during my
actual titration. We found that weak acids complicate calculating pH at equivalence point because you
need to use and ice chart which is not needed when working with strong acids and bases. By looking at
the graph it can help you determine both your Equivalence point and your half equivalence point which
can help determine the pKa. The volume of titrant added halfway up the jump is roughly your
equivalence point. When that volume is divided in half you have your half equivalence point. When
looking at the graph whichever pH corresponds to the volume at the half equivalence point is the pKa.
This is helpful when trying to determine the identity of the unknown acid because you have no molar
mass or Ka to work with.